The area bounded by the curve $y = f(x)$,the coordinate axes,and the line $x = x_1$ is given by $x_1 \cdot e^{x_1}$. Therefore,$f(x)$ equals:

Find the area of the region bounded by the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$. (in $\pi$)

The area bounded by the $x-$ axis,the curve $y = f(x)$ and the lines $x = 1$ and $x = b$ is equal to $\sqrt{b^2 + 1} - \sqrt{2}$ for all $b > 1$. Then $f(x)$ is:

Find the area enclosed by the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.

The area of the region bounded by the curve $5y = 5 - x$,the $X$-axis,and the lines $x = 1$ and $x = 4$ is . . . . . . sq. units.

Let $S(\alpha) = \{(x,y) : y^2 \leq x, 0 \leq x \leq \alpha\}$ and $A(\alpha)$ be the area of the region $S(\alpha)$. If for a $\lambda, 0 < \lambda < 4, A(\lambda) : A(4) = 2 : 5$,then $\lambda$ equals: